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FP3 - 2 Limits Questions - Stuck

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    Q1:

    \displaystyle\lim_{x\to \infty} \dfrac{x+e^x}{x-e^x}

    I have tried multiplying top and bottom by  x + e^x and that didn;t get me anywhere, I then tried expanding the ex's and after some working eventually got stuck at

    \displaystyle \dfrac{1+2x+\dfrac{x^2}{2}+...}{-1-\frac{x^2}{2}+\cdots}

\

= (-1)(1+2x+\dfrac{x^2}{2} + \cdots)(1+ (\dfrac{x^2}{2} + \dfrac{x^3}{6}+\cdots))^{-1}

    I have been on it for a while so a solution would be better than a hint, unless there is something small that I am missing or a mistake somewhere.



    Q2:

     \displaystyle F(x) = \dfrac{e^x}{1-x}

\

\text{Show that} F(x) \to -\infty \ \text {as} \ x \to \infty

    For this one I expanded ex and brought the (1-x) to the top and expanded it, multiplied it out and I am stuck at

    (1+2x+\frac{5x^2}{2} \cdots)

    Same as above, all help is appreciated.

    Thnx
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    On Q1 you could try multiplying top and bottom by e^-x.
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    (Original post by ian.slater)
    On Q1 you could try multiplying top and bottom by e^-x.
    Yea, that works lol. The answer is -1, why didn't I see something so simple !!
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    (Original post by member910132)
    Ready To Answer Now - Latex Problem Sorted

    Q1:

    \displaystyle\lim_{x\to \infty} \dfrac{x+e^x}{x-e^x}

    I have tried multiplying top and bottom by  x + e^x and that didn;t get me anywhere, I then tried expanding the ex's and after some working eventually got stuck at

    \displaystyle \dfrac{1+2x+\dfrac{x^2}{2}+...}{-1-\frac{x^2}{2}+\cdots}

\

= (-1)(1+2x+\dfrac{x^2}{2} + \cdots)(1+ (\dfrac{x^2}{2} + \dfrac{x^3}{6}+\cdots))^{-1}

    I have been on it for a while so a solution would be better than a hint, unless there is something small that I am missing or a mistake somewhere.



    Q2:

     \displaystyle F(x) = \dfrac{e^x}{1-x}

\

\text{Show that} F(x) \to -\infty \ \text {as} \ x \to \infty

    For this one I expanded ex and brought the (1-x) to the top and expanded it, multiplied it out and I am stuck at

    (1+2x+\frac{5x^2}{2} \cdots)

    Same as above, all help is appreciated.

    Thnx
    second one:
    large +ve x makes e^x +ve and 1-x -ve so F is -ve. Now consider the relative order of exponentials and linear expressions in x.
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    (Original post by ian.slater)
    On Q1 you could try multiplying top and bottom by e^-x.
    Just tried that for Q2 and I get:

    \displaystyle\dfrac{1}{e^{-x}-xe^{-x}}Using the general result that xe^(-x) tends to 0 as x tends to infinity my only problem is showing how F(x) goes to negative infinity and not just infinity.
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    (Original post by member910132)
    Ready To Answer Now - Latex Problem Sorted

    Q1:

    \displaystyle\lim_{x\to \infty} \dfrac{x+e^x}{x-e^x}

    I have tried multiplying top and bottom by  x + e^x and that didn;t get me anywhere, I then tried expanding the ex's and after some working eventually got stuck at

    \displaystyle \dfrac{1+2x+\dfrac{x^2}{2}+...}{-1-\frac{x^2}{2}+\cdots}

\

= (-1)(1+2x+\dfrac{x^2}{2} + \cdots)(1+ (\dfrac{x^2}{2} + \dfrac{x^3}{6}+\cdots))^{-1}

    I have been on it for a while so a solution would be better than a hint, unless there is something small that I am missing or a mistake somewhere.



    Q2:

     \displaystyle F(x) = \dfrac{e^x}{1-x}

\

\text{Show that} F(x) \to -\infty \ \text {as} \ x \to \infty

    For this one I expanded ex and brought the (1-x) to the top and expanded it, multiplied it out and I am stuck at

    (1+2x+\frac{5x^2}{2} \cdots)

    Same as above, all help is appreciated.

    Thnx
    You could use L'Hopital' s Theorem...they both come out very easily then.....
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    (Original post by ben-smith)
    second one:
    large +ve x makes e^x +ve and 1-x -ve so F is -ve. Now consider the relative order of exponentials and linear expressions in x.
    Sorry, I don't follow the second part.
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    (Original post by mikelbird)
    You could use L'Hopital' s Theorem...they both come out very easily then.....
    That isn't on the AQA FP3 syllabus and I don't want to learn stuff that ain't on the syllabus.
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    (Original post by member910132)
    Sorry, I don't follow the second part.
    from an intuitive point of view, e^x is like a polynomial of very high degree and so it's going to tend to infinity faster than a linear function.
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    (Original post by member910132)
    Just tried that for Q2 and I get:

    \displaystyle\dfrac{1}{e^{-x}-xe^{-x}}Using the general result that xe^(-x) tends to 0 as x tends to infinity my only problem is showing how F(x) goes to negative infinity and not just infinity.
    x e^-x > e^-x
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    For Q2 I am almost there, I just need to show that  \dfrac{1}{e^{-x} - xe^{-x}} will tendo to  \displaystyle -\infty.  \displaystyle e^{-x}\to 0, xe^{-x} \to 0 as  \displaystyle x\to \infty but I need to show that they tend to  \displaystyle  0 from the negative as only then will the function  \displaystyle \to -\infty.
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    (Original post by mikelbird)
    You could use L'Hopital' s Theorem...they both come out very easily then.....
    Not allowed.
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    So is it sufficient for me to say that  \displaystyle  \dfrac{1}{e^{-x} - xe^{-x}} \to -\infty as  x\to \infty because both terms in the denominator tend to zero and  xe^{-x} > e^x ?

    Edit: So can it be said that as  x\to \infty, \ e^{-x} - xe^{-x}\to 0- and hence  \displaystyle  \dfrac{1}{e^{-x} - xe^{-x}} \to -\infty?
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    Bump - will anyone verify my above post so I can finish this thread ?
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    (Original post by member910132)
    Bump - will anyone verify my above post so I can finish this thread ?
    I'd be happy with it.
 
 
 
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